Answer
$0.024in^3/min$
Work Step by Step
Step 1. Identify the given quantities: $h=6in$, $\frac{dr}{dt}=\frac{1}{1000}in/3min=\frac{1}{3000}in/min$, $2r=3.8in, r=1.9in$
Step 2. Volume: $V=\pi r^2h=6\pi r^2$, we have $\frac{dV}{dt}=6\pi (2r)\frac{dr}{dt}=12\pi(1.9)(\frac{1}{3000})\approx0.024in^3/min$
